In a certain chemical process a substance A reacts with and reduces a substance B. The masses of A and B at time t after the start of the process are x and y respectively. It is given that dy/dt = -0.2xy and x = 10 / (1 + t)². At the beginning of the process y = 100.

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

1. [B1] State dy/dt = -2y / (1+t)² , or equivalent
2. [M1] Separate variables correctly and attempt integration of one side
3. [A1] Obtain term ln y, or equivalent
4. [A1] Obtain term 2 / (1+t), or equivalent
5. [M1] Use y = 100 and t = 0 to evaluate a constant, or as limits in an expression containing terms of the form a ln y and b/(1+t)
6. [A1] Obtain correct solution in any form, e.g. ln y = 2/(1+t) - 2 + ln 100
7. [B1] State that the mass of B approaches 100/e² , or exact equivalent
8. [B1] State or imply that the mass of A tends to zero